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Homework Help: Calculating a real integral with a complex integral

  1. Dec 7, 2013 #1
    1. The problem statement, all variables and given/known data
    Evaluate the following real integral using complex integrals:
    [itex]\int_0^\infty \frac{cos(2x)}{x^2+4}dx[/itex]

    2. Relevant equations
    Cauchy's Residue Theorem for simple pole at a: [itex]Res(f;a)=\displaystyle\lim_{z\rightarrow a} (z-a)f(z)[/itex]

    3. The attempt at a solution
    Since the function [itex]\frac{cos(2x)}{x^2+4}dx[/itex] is even, [itex]\int_0^\infty \frac{cos(2x)}{x^2+4}dx =\frac{1}{2} \int_{-\infty}^\infty \frac{cos(2x)}{x^2+4}dx[/itex]

    Consider the real part of [itex]f(z)=\frac{e^{2iz}}{z^2+4}[/itex]

    [itex]\int_0^\infty \frac{cos(2x)}{x^2+4}dx[/itex]=[itex]\frac{1}{2}Re(\displaystyle\lim_{R\rightarrow \infty}\int_{[-R,R]} \frac{e^{2iz}}{z^2+4})dz[/itex]

    We have [itex]\int_{[-R,R]}f(z)dz=\int_\gamma f(z)-\int_{\gamma R} f(z)dz[/itex]

    where [itex]\int_{\gamma R} f(z)dz[/itex] tends to zero as can be shown by the estimation theorem.

    To calculate [itex]\int_\gamma f(z)[/itex] we find the residue of f at the simple pole z=2i:

    [itex]Res(f;2i)=\lim_{z\rightarrow {2i}}(z-2i)\frac{e^{2iz}}{z^2+4}=\lim_{z\rightarrow {2i}}\frac{e^{2iz}+2ie^{2iz}(z-2i)}{2z}=\frac{1}{4ie^4}[/itex]

    This is obviously incorrect since there can be no imaginary part
    Last edited: Dec 7, 2013
  2. jcsd
  3. Dec 7, 2013 #2


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    Homework Helper

    The integral is [itex]2 \pi i[/itex] times the residue.
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